Equivalence between expected parameter increments in "Off-Policy Temporal-Difference Learning with Function Approximation"
First part is correct
\begin{align}
&amp;\sum_{n=1}^{\infty} \alpha(1-\lambda)\lambda^{n-1} (\bar R_t^{(n)} - \theta^T \phi_t)\\
=&amp; \alpha[\sum_{n=1}^{\infty} (1-\lambda)\lambda^{n-1} \bar R_t^{(n)} - \sum_{n=1}^{\infty} (1-\lambda)\lambda^{n-1} \theta^T \phi_t]
\end{align}
$\sum_{n=1}^{\infty} (1-\lambda)\lambda^{(n-1)}$ sums to $1$ so we have
\begin{equation}
\alpha[\sum_{n=1}^{\infty} (1-\lambda)\lambda^{n-1} \bar R_t^{(n)} - \theta^T \phi_t]
\end{equation}
For the second part it's enough to prove equivalence for any $n$ because result contains sum over $n$. If you have 2 sums $\sum x_n$, $\sum y_n$ then the sums will be equal if for any $n$, $x_n = y_n$.  

For the third part, we are in state $s_t$ and we already took action $a_t$ so we have
\begin{align}
&amp;E_b \{ \sum_{t=0}^{\infty} (\bar R_t^{(n)} - \theta^T\phi_t)\phi_t \rho_1\rho_2\cdots\rho_t\}\\
=&amp; \sum_{t=0}^{\infty} E_b \{(\bar R_t^{(n)} - \theta^T\phi_t)\phi_t \rho_1\rho_2\cdots\rho_t\}\\
=&amp; \sum_{t=0}^{\infty} E_b \{\phi_t \rho_1\rho_2\cdots \rho_t\} E_b \{(\bar R_t^{(n)} - \theta^T\phi_t)|s_t, a_t\}
\end{align}
that is because $\rho_i, i = 1, \ldots, t-1$ depends on $s_i, a_i$. Because of Markov property expectation over $\bar R_t^{(n)}$ doesn't depend on those state it only depends on $s_t, a_t$ so they are independent. We don't need to consider $\phi_t$ and $\rho_t$ in expectation over $\bar R_t^{(n)}$ either because, like I said, we are in state $s_t$ and we took $a_t$ so they are already decided they would be considered a constant. We can then split total expectation in part $E_b \{\phi_t \rho_1\rho_2\cdots \rho_t\}$ for getting to state $s_t$ and taking action $a_t$ and part $E_b \{(\bar R_t^{(n)} - \theta^T\phi_t)|s_t, a_t\}$ for expectation over $\bar R_t^{(n)}$ after we got to state $s_t$ and took action $a_t$.